$A = \begin{pmatrix} -3 & 7 & -10 \ 5 & 2 & 8 \ -9 & 0 & -6 \ 4 & -2 & 1 \ 7 & -8 & 9 \ -1 & 3 & -5 \ 10 & -4 & 6 \end{pmatrix}$ $B = \begin{pmatrix} 2 & -3 & 4 \ 1 & 5 & -6 \ -7 & 8 & 9 \end{pmatrix}$ $C = \begin{pmatrix} 1 & 5 & 0 & 0 \ 0 & 0 & 8 & 0 \ 1 & 1 & 0 & -4 \end{pmatrix}$ $D = \begin{pmatrix} 2 & -4 & 1 & -3 \ -4 & 3 & -2 & 4 \ 0 & 0 & 5 & 2 \ 0 & 1 & 2 & -6 \end{pmatrix}$
Added by William J.
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Given A = [6 0; 4 3; 1 2], the transpose of A (AT) is: AT = [6 4 1; 0 3 2] Now, let's calculate AT A: AT A = [6 4 1; 0 3 2] * [6 0; 4 3; 1 2] = [6*6 + 4*4 + 1*1 6*0 + 4*3 + 1*2; 0*6 + 3*4 + 2*1 0*0 + 3*3 + 2*2;] = [36 + 16 + 1 0 + 12 + 2; 0 + 12 + 2 Show more…
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